Prove $\sum \frac{1}{(f(n))^2}$ converges.

Question

Let $f$ be an unbounded, non-decreasing function, With the following property:

For any positive sequence $a_n$ s.t. $\sum a_n$ converges, $\sum\frac{1}{f(\frac{1}{a_n})}$ also converges.

Prove, Or disprove:

$\displaystyle\sum \frac{1}{(f(n))^2}$ also converges.

So far, I have tried to show convergence of $\sum \frac{1}{(f(n))^2}$ by showing the limit of $\frac{f(n)^2}{f(1/a_n)}$ exists but it got me no where and after a couple of attempts I realized that's not the way to go.

Any attempts to find an upper bound failed as well.

Any hints will be appreciated.

Answer 1

Let's apply the hypothesis with $a_n = \dfrac{1}{n^2}$. It gives that the series $\displaystyle{\sum \dfrac{1}{f(n^2)}}$ converges.

Let $b_n = \dfrac{1}{f(n)}$, so that we know that $\displaystyle{\sum b_{n^2}}$ converges.

For every $n \geq 1$, using the fact that $(b_n)$ decreases, one has $$\sum_{k=1}^{n^2-1} b_k^2 = \sum_{k=1}^{n-1} \sum_{j=k^2}^{(k+1)^2-1} b_j^2 \leq \sum_{k=1}^{n-1} (2k+1)(b_{k^2})^2$$

But because $\displaystyle{\sum b_{n^2}}$ converges and $(b_{n^2})$ decreases, we get by this very classical fact that $kb_{k^2}$ tends to $0$, hence $(2k+1)b_{k^2}$ tends to $0$, hence is bounded, hence the RHS of the above inequality is a convergent series.

By comparison, the series $\displaystyle{\sum b_k^2}$ converges, i.e. the series $\displaystyle{\sum \dfrac{1}{f(n)^2}}$ converges.

Answer 2

If $f(x)>rx$ for some positive $r$ and all sufficiently large $x$, then the answer's obvious.

Therefore, try showing that $\displaystyle\liminf_{x\to\infty}\dfrac{f(x)}{x}$ must be greater than $0$.

Hint: suppose that $f(x)=x\cdot z(x)$, where $\displaystyle\liminf_{x\to\infty}z(x)=0$.

Construct a positive sequence $a_n$ such that $\displaystyle\sum_{n=1}^\infty a_n$ converges but $\displaystyle\sum_{n=1}^\infty a_n\dfrac{1}{z\left(\frac{1}{a_n}\right)}$ diverges.

If you want to read the proof

This is equivalent to the following:

Suppose that $j(x)$ is positive and $\displaystyle\limsup_{x\to0^+}j(x)=\infty$. Note, I'm talking about a limit from the right. Construct a sequence $a_n$ such that $\displaystyle\sum_{n=1}^\infty a_n$ converges but $\displaystyle\sum_{n=1}^\infty a_n j(a_n)$ diverges.

Solution: what does the condition on $j(x)$ imply? It implies that there exists a strictly decreasing positive sequence $b_n\to0$ such that $j(b_n)$ strictly increases to $\infty$. We are going to exploit the existence of such $b_n$ to create our $a_n$. Our sequence $a_n$ shall be made by selecting some of the $b_n$ and repeating each of these a certain number of times in a judicious way.

Let $n_k$ be the smallest natural number $n$ such that $b_n<\dfrac{1}{k^2}$ but $j(b_n)>k$. The $k^\text{th}$ block in our sequence shall consist in $b_{n_k}$ repeated $t(k)$ times, where $t(k)$ is the smallest natural number $t$ such that $\dfrac{1}{k^2}<t\;b_{n_k}<\dfrac{2}{k^2}$.

Now,

$\displaystyle\sum_{n=1}^\infty a_n=\sum_{k=1}^\infty t(k)b_{n_k}\leq\sum_{k=1}^\infty \dfrac{2}{k^2}$ therefore converges

but

$\displaystyle\sum_{n=1}^\infty a_n j(a_n)=\sum_{k=1}^\infty t(k)b_{n_k}j(b_{n_k})\geq\sum_{k=1}^\infty \dfrac{k}{k^2}$ therefore diverges.

Answer 3

First, observe that the series $\displaystyle\sum \frac{1}{n^2}$ converges, so $\displaystyle\sum \frac{1}{f(n^2)}$ does, and second, for all $n\in\mathbb{N}$,$\displaystyle\sum_{k=n^2+1}^{(n+1)^2}\frac{1}{f(k)^2}\leq \frac{2n+1}{f(n^2)}\frac{1}{f(n^2)}$ follows by the monoticity of $f$. Thus, we may proceed if we can show that the sequence $\displaystyle\big (\frac{2n+1}{f(n^2)}\big )$ is bounded.

Suppose not. Then, for any $M>0$, we could find $n\in\mathbb{N}$ such that $\displaystyle\frac{n}{f(n^2)}\geq M$, which implies (again by the monotonicity of $f$) that $\displaystyle\sum_{k=1}^n \frac{1}{f(k^2)}\geq \frac{nM}{n}$, contradicting the convergence of the series. Thus, let $M$ bound the sequence $\big (\displaystyle\frac{2n+1}{f(n^2)}\big )$ instead.

Then, for any $N\in\mathbb{N}$, $$\sum_{n=1}^{N^2}\frac{1}{f(n)^2}=\sum_{n=1}^{N-1}\sum_{k=n^2+1}^{(n+1)^2}\frac{1}{f(k)^2}+\frac{1}{f(1)}\leq\sum_{n=1}^{N-1}\frac{2n+1}{f(n^2)}\frac{1}{f(n^2)}+\frac{1}{f(1)}\leq M\sum \frac{1}{f(n^2)}+\frac{1}{f(1)},$$ so the result follows.